# Feynman rules out of the Lagrangian

Accordingly to chapter 10, section 10.6 Feynman Rules of 'Introduction to Elementary Particles' by David Griffiths, there is a way to extract the vertex and propagators just by inspection of the Lagrangian:

1) Propagators: take the Euler-Lagrange equations of the free fields and the inverse of the operators in momentum space that act on the fields and multiplied by $$i$$ are the propagators for each one.

2) Vertex: take interaction Lagrangian and multiply it by $$i$$. Make use of the prescription $$i\partial_\mu \rightarrow k_\mu$$ and rub out the fields. The remaining is the vertex.

My questions are:

a) If I want to compute the vertex for the interaction Lagrangian $${\cal L}_{int} = g\varphi\partial_\mu \varphi \partial^\mu\varphi$$ (all scalar fields), by rule 2) I get $${\rm vertex} = -igk_1k_2$$. Nevertheless, apparently the solution is $$-2ig(k_1k_2 + k_1k_3 + k_2k_3)$$. Where did these extra factors come out?

b) If I had a similar interaction as in a) but changing one field for a new one $$\chi$$ (scalar too), so $${\cal L}_{int} = g\chi \partial_\mu \varphi \partial^\mu\varphi$$, would the solution be $$-2igk_1k_2$$ with $$k_i$$ the momenta of $$\varphi$$ fields?

c) This book grants you that for QCD, with $${\cal L}_{int}^{3\ fields} = g\{[\partial^\mu A^\nu - \partial^\nu A^\mu]·(A_\mu \times A_\nu) + (A^\mu \times A^\nu)·[\partial_\mu A_\nu - \partial_\nu A_\mu]\}$$, you can obtain the 3 fields vertex known as $${\rm vertex} = -gf^{\alpha \beta \gamma}[g_{\mu \nu}(k_1 - k_2)_\lambda + g_{\nu \lambda}(k_2 - k_3)_\mu + g_{\lambda \mu}(k_3 - k_1)_\nu]$$. I've tried to get it from rule 2) but I wasn't able to.

• a) Notice that you arbitrarily chose $k_1 k_2$ while you just have well could have chosen $k_1 k_3$ or $k_2 k__1$, etc. There are six ways to choose two momenta out of three, so you end up with six terms. Then you notice that $k_i k_j = k_j k_i$ so you can condense it into three terms times 2. Srednicki’s book does a good job of explaining Feynman rules and is free online. – Diffycue Nov 8 '18 at 1:55
• Yes, I know. I'm starting to work with it (case a) is one of the non-solved exercises of that book). Are the rules that I talk about explained in this book? I don't see it – Vicky Nov 8 '18 at 2:03
• the Feynman rules are explained in the chapter on Feynman rules – Diffycue Nov 8 '18 at 18:34
• I meant this tricks/rules for Feynman rules – Vicky Nov 8 '18 at 18:39

Try to evaluate (e.g. for the $$g\phi \partial_\mu \phi \partial^{\mu} \phi$$ interaction) the $$\mathcal{O}(g)$$ greens functions $$G(x_1,x_2,x_3)$$ of 3 external scalar particles, e.g. by using wicks theorem, and see what the vertex rule is. In position space you get 6 terms of the form: $$-ig\int d^4x D(x_1-x) (\partial_\mu D)(x_2-x) (\partial^\mu D)(x_3-x)$$. Then consider the momentum space greensfunction $$\tilde{G}(k_1,k_2,k_3) = \int d^4x_1 \int d^4x_2 \int d^4x_3 G(x_1,x_2,x_3) e^{-ik_1x_1}e^{-ik_2x_2}e^{-ik_3x_3}$$ (usual convention is all momenta ingoing, i.e. $$e^{-ikx}$$). This gives you the $$g k_2 k_3$$ term. When you do this for all the other contractions you get $$3$$ non equal contributions and $$2$$ each of the total $$6$$ are equal (therefore the factor $$2$$).
The QCD case is also the same, however all $$6$$ are non equal, because they you have an additional space time and colour index. Therefore you get the $$6$$ terms.
• So, under your way, for b) case, I could make $D(x_1 - x) \rightarrow D^\chi(x_1 - x)$ (propagator for $\chi$) and therefore there will be just two possible contractions by switching $x_2$ and $x_3$, so my solution will be the right one, true? – Vicky Nov 8 '18 at 0:05
• yes. You are correct. The two contractions are the same and you get $-2igk_2k_3$. – jkb1603 Nov 8 '18 at 1:23